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I have a problem with what looks like a very easy equation to solve $\left|\frac{1 + z}{1- i\overline z}\right| = 1$ . ($z$ is a complex number, $\overline z$ is a conjugate of $z$) I got stuck at the point when after replacing $\overline z = a-bi $ and $z = a +bi$ and getting rid of absolute value I end up with $a^2 +a-b =0$. I have no idea how to follow this up or wether I should take totally different approach from the begining. I'd be very gratefull if someone could guide me into the right solution.

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You want $|1+z|=|1-i\bar z|$. The left side here is the distance from $z$ to $-1$. The right side equals $|i+\bar z|$ which in turn equals $|-i+z|$, the distance from $z$ to $i$. So $z$ satisfies your equation iff it is equidistant from $-1$ and $i$. These $z$'s form a straight line in the complex plane, whose equation you can find by drawing a picture.

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Something went wrong in your evaluation. After the substitution $z=a+ib$, you should have $$(1+a)^2+b^2=|1+a+ib|^2=|1-b-ia|^2=(1-b)^2+(-a)^2$$ that is, after a few simplifications, $a=-b$.

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Since $|1+z|=|1-i\overline{z}|$ and $|w| = |\overline{w}|$, you can rewrite like this $$|z-(-1)|=|1+z|=|1-i\overline{z}| = |\overline{1-i\overline{z}}| =|1+iz| =|i||-i+z| = |z-i| $$

So $z$ is at equal distance from $-1$ and $i$. So $z$ is on perpendicular bisector of segment between $-1$ and $i$.

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Hint: Let $z=a+ib$, then $|\frac{1+z}{1-i\overline{z}}|=\frac{|1+z|}{|1-i\overline{z}|}=\frac{|(1+a)+ib|}{|(1-b)-ia|}=\frac{\sqrt{(1+a)^2+b^2}}{\sqrt{(1-b)^2+a^2}}=\frac{\sqrt{1+a^2+2a+b^2}}{\sqrt{1+b^2-2b+a^2}}=1$

Now this equality is true if and only if $a=-b$ which gives us that if $z=a+ib$ then $iz=ai-b=a-bi=\overline{z}$

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Take this $$|1+z|=|1-i\bar{z}|$$ then square this babies $$(1+z)(1+\bar{z})=(1-i\bar{z})(1+iz)$$ $$1+z+\bar{z}+z\bar{z}=1-i\bar{z}+iz+z\bar{z} $$ so $$ z+\bar{z}=i(z-\bar{z})$$ $$2Re(z)=i(i2Im(z)$$ therefore $$Re(z)=-Im(z)$$

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Let $z=x+yi$ and then from $$ |1+z|=|1-i\bar{z}| $$ it is easy to obtain $$ (x+1)^2+y^2=(-y+1)^2+x^2, $$ So $$ x=-y $$ and hence $z=(1-i)x$.

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